Write a program to find the node at which the intersection of two singly linked lists begins.

For example, the following two linked lists:

A:          a1 → a2                   ↘                     c1 → c2 → c3                   ↗            B:     b1 → b2 → b3

begin to intersect at node c1.

Notes:

• If the two linked lists have no intersection at all, return null.
• The linked lists must retain their original structure after the function returns.
• You may assume there are no cycles anywhere in the entire linked structure.
• Your code should preferably run in O(n) time and use only O(1) memory.

想法：逐个判断，如果p1==p2,那么返回其中一个即可，如果在比较的过程中，一个链表达到尾端，那么将其移动到另外一个链表到的位置，重新开始比较。即加入headB到达节点p2,链表headA到了终点，那么将headA移动到p2，重新开始比较。

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     struct ListNode *next; * }; */struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) {    if(headA == NULL || headB == NULL){        return NULL;    }    struct ListNode* p1 = headA;    struct ListNode* p2 = headB;    while(p1 != NULL && p2!= NULL && p1 != p2){        p1 = p1->next;        p2 = p2->next;                if(p1 == p2) return p1;        if (p1 == NULL) p1 = headB;        if (p2 == NULL) p2 = headA;    }    return p1;}